Closed form of a sum with a falling factorial

Question

I am having trouble finding the closed form for this:

$\sum_{k=1}^n k^\underline{-2} + 2^k$

If it were just

$\sum_{k=1}^n 2^k + 2k$

I would just use the geometric series to solve for the closed form, but I do not know how to apply the geometric sum to the falling factorial. Can someone help me out?

Answer 1

Using the definition of Falling Factorial, rewrite the sum as $$\begin{align} \sum_{k=1}^n k^\underline{-2} + 2^k &= \sum_{k=1}^n k(k-1) + 2^k \\&= \sum_{k=1}^n k^2-k + 2^k \\&= \sum_{k=1}^nk^2 - \sum_{k=1}^nk + \sum_{k=1}^n 2^k \end{align}$$ Then, use the known formulas.